Pertanyaan

Jika x → 2 1 lim ⎝ ⎛ x − 2 1 3 a x 3 + b − 2 ⎠ ⎞ = A , maka nilai x → 2 1 lim ⎝ ⎛ 4 x 2 − 1 3 8 a x 3 + 8 b − 2 x ⎠ ⎞ = ....

Jika $x \to \frac{1}{2} lim ⎝ ⎛ \frac{3 a x ^{3} + b - 2}{x - \frac{1}{2}} ⎠ ⎞ = A$ , maka nilai $x \to \frac{1}{2} lim ⎝ ⎛ \frac{3 \frac{a x ^{3}}{8} + \frac{b}{8} - 2 x}{4 x ^{2} - 1} ⎠ ⎞ = ....$

Ikuti Tryout SNBT & Menangkan E-Wallet 100rb

Habis dalam

Klaim

A. Allamah

Master Teacher

Mahasiswa/Alumni Universitas Brawijaya

Jawaban terverifikasi

Jawaban

ka , maka nilai

ka $limit as x rightwards arrow 1 half of open parentheses fraction numerator cube root of a x cubed plus b end root minus 2 over denominator x minus begin display style 1 half end style end fraction close parentheses equals A$ , maka nilai $limit as x rightwards arrow 1 half of open parentheses fraction numerator cube root of begin display style fraction numerator a x cubed over denominator 8 end fraction plus b over 8 end style end root minus 2 x over denominator 4 x squared minus begin display style 1 end style end fraction close parentheses equals fraction numerator A minus 4 over denominator 8 end fraction$

Pembahasan

Ingat Definisi turunan, jika Penyelesaian dapat menggunakan limit fungsi dengan metode L'Hospital dan menggunakan turunan Jika limit fungsi berbentuk , maka limit fungsi tersebut bisa diselesaikan dengan menggunakan turunan, yaitu Menggunakan sifat limit berikut Serta memisalkan , didapatkan Dengan demikian, dengan dalil L'Hospital diperoleh Akan dicari nilai dari . Dengan menjabarkan bentuk aljabar sebagai berikut Akan dicari nilai dengan L'Hospital. Jadi,nilai dari . Dengan demikian,ka , maka nilai

Ingat Definisi turunan, jika

table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell equals cell a x to the power of n rightwards arrow f apostrophe open parentheses x close parentheses equals a n x to the power of n minus 1 end exponent end cell end table

Penyelesaian dapat menggunakan limit fungsi dengan metode L'Hospital dan menggunakan turunan

Jika limit fungsi berbentuk $limit as x rightwards arrow k of fraction numerator f open parentheses x close parentheses over denominator g open parentheses x close parentheses end fraction equals 0 over 0$ , maka limit fungsi tersebut bisa diselesaikan dengan menggunakan turunan, yaitu

$limit as x rightwards arrow k of fraction numerator f open parentheses x close parentheses over denominator g open parentheses x close parentheses end fraction equals limit as x rightwards arrow k of fraction numerator f apostrophe open parentheses x close parentheses over denominator g apostrophe open parentheses x close parentheses end fraction$

Menggunakan sifat limit berikut

$table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow k of f open parentheses x close parentheses plus-or-minus g open parentheses x close parentheses end cell equals cell limit as x rightwards arrow k of f open parentheses x close parentheses plus-or-minus limit as x rightwards arrow k of g open parentheses x close parentheses end cell row cell limit as x rightwards arrow k of fraction numerator f open parentheses x close parentheses over denominator g open parentheses x close parentheses end fraction end cell equals cell fraction numerator limit as x rightwards arrow k of f open parentheses x close parentheses over denominator limit as x rightwards arrow k of g open parentheses x close parentheses end fraction end cell row cell limit as x rightwards arrow k of a f open parentheses x close parentheses end cell equals cell a limit as x rightwards arrow k of f open parentheses x close parentheses end cell end table$

Serta memisalkan f open parentheses x close parentheses equals cube root of a x cubed plus b end root , didapatkan

$table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 half of open parentheses fraction numerator f open parentheses x close parentheses minus 2 over denominator x minus begin display style 1 half end style end fraction close parentheses end cell equals cell limit as x rightwards arrow 1 half of open parentheses fraction numerator f apostrophe open parentheses x close parentheses over denominator 1 end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 1 half of f apostrophe open parentheses x close parentheses end cell row blank equals cell limit as x rightwards arrow 1 half of f apostrophe open parentheses x close parentheses end cell end table$

Dengan demikian, dengan dalil L'Hospital diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row blank blank blank row cell limit as x rightwards arrow 1 half of f apostrophe open parentheses x close parentheses end cell equals A end table

Akan dicari nilai dari $limit as x rightwards arrow 1 half of open parentheses fraction numerator cube root of begin display style fraction numerator a x cubed over denominator 8 end fraction plus b over 8 end style end root minus 2 x over denominator 4 x squared minus begin display style 1 end style end fraction close parentheses$ . Dengan menjabarkan bentuk aljabar sebagai berikut

$table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator cube root of begin display style fraction numerator a x cubed over denominator 8 end fraction plus b over 8 end style end root minus 2 x over denominator 4 x squared minus begin display style 1 end style end fraction end cell equals cell fraction numerator cube root of begin display style 1 over 8 open parentheses a x cubed plus b close parentheses end style end root minus 2 x over denominator 4 x squared minus begin display style 1 end style end fraction end cell row blank equals cell fraction numerator cube root of begin display style 1 over 8 end style end root times cube root of open parentheses a x cubed plus b close parentheses end root minus 2 x over denominator 4 x squared minus 1 end fraction end cell row blank equals cell fraction numerator begin display style 1 half end style cube root of open parentheses a x cubed plus b close parentheses end root minus 2 x over denominator 4 x squared minus 1 end fraction end cell end table$

Akan dicari nilai $limit as x rightwards arrow 1 half of open parentheses fraction numerator cube root of begin display style fraction numerator a x cubed over denominator 8 end fraction plus b over 8 end style end root minus 2 x over denominator 4 x squared minus begin display style 1 end style end fraction close parentheses$ dengan L'Hospital.

$table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 1 half of open parentheses fraction numerator cube root of begin display style fraction numerator a x cubed over denominator 8 end fraction plus b over 8 end style end root minus 2 x over denominator 4 x squared minus begin display style 1 end style end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 1 half of fraction numerator begin display style 1 half cube root of open parentheses a x cubed plus b close parentheses end root minus 2 x end style over denominator 4 x squared minus 1 end fraction end cell row blank equals cell limit as x rightwards arrow 1 half of fraction numerator begin display style 1 half f open parentheses x close parentheses minus 2 x end style over denominator 4 x squared minus 1 end fraction space end cell row blank equals cell limit as x rightwards arrow 1 half of fraction numerator begin display style 1 half f apostrophe open parentheses x close parentheses minus 2 end style over denominator 8 x end fraction space end cell row blank equals cell fraction numerator begin display style limit as x rightwards arrow 1 half of 1 half f apostrophe open parentheses x close parentheses minus limit as x rightwards arrow 1 half of 2 end style over denominator limit as x rightwards arrow 1 half of 8 x end fraction end cell row blank equals cell fraction numerator begin display style 1 half limit as x rightwards arrow 1 half of f apostrophe open parentheses x close parentheses minus 2 end style over denominator 8 limit as x rightwards arrow 1 half of x end fraction end cell row blank equals cell fraction numerator begin display style 1 half end style A minus begin display style 2 end style over denominator 8 times begin display style 1 half end style end fraction end cell row blank equals cell fraction numerator begin display style 1 half open parentheses A minus 4 close parentheses end style over denominator 4 end fraction end cell row blank equals cell fraction numerator begin display style A minus 4 end style over denominator 8 end fraction end cell end table$

Jadi, nilai dari $limit as x rightwards arrow 1 half of open parentheses fraction numerator cube root of begin display style fraction numerator a x cubed over denominator 8 end fraction plus b over 8 end style end root minus 2 x over denominator 4 x squared minus begin display style 1 end style end fraction close parentheses equals fraction numerator A minus 4 over denominator 8 end fraction$ .

Dengan demikian, ka $limit as x rightwards arrow 1 half of open parentheses fraction numerator cube root of a x cubed plus b end root minus 2 over denominator x minus begin display style 1 half end style end fraction close parentheses equals A$ , maka nilai $limit as x rightwards arrow 1 half of open parentheses fraction numerator cube root of begin display style fraction numerator a x cubed over denominator 8 end fraction plus b over 8 end style end root minus 2 x over denominator 4 x squared minus begin display style 1 end style end fraction close parentheses equals fraction numerator A minus 4 over denominator 8 end fraction$

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

4.1 (10 rating)

Dustin Nathanael

Ini yang aku cari!

Amanda Deby febrina

Makasih ❤️

A. Sitti Nurhalijah

Jawaban tidak sesuai

Dani

Jawaban tidak sesuai

Pertanyaan serupa

Diberikan x → a lim f ( x ) = 3 dan x → a lim g ( x ) = − 1 . Hitunglah nilai setiap limit berikut. a. x → a lim f 2 ( x ) + g 2 ( x ) b. x → a lim f ( x ) + g ( x ) 2 f ( x ) − 3 g...

3.7

Jawaban terverifikasi

Diberikan x → a lim f ( x ) = 3 dan x → a lim g ( x ) = − 1 . Hitunglah nilai setiap limit berikut. a. x → a lim f 2 ( x ) + g 2 ( x ) b. x → a lim f ( x ) + g ( x ) 2 f ( x ) − 3 g...

3.7

Jawaban terverifikasi

Diketahui x → a lim f ( x ) = 8 , x → a lim g ( x ) = − 3 , dan x → a lim h ( x ) = 4 . Tentukan a. x → a lim ( 2 f ( x ) + 3 g ( x ) − h ( x ) ) .

4.7

Jawaban terverifikasi

Diketahui x → a lim f ( x ) = 8 , x → a lim g ( x ) = − 3 , dan x → a lim h ( x ) = 4 . Tentukan a. x → a lim ( 2 f ( x ) + 3 g ( x ) − h ( x ) ) .

4.7

Jawaban terverifikasi

Diketahui x → 2 lim f ( x ) = 3 dan nilai limit berikut. ( i ) x → 2 lim [ f ( x ) − f 2 ( x ) ] = 6 ( ii ) x → 2 lim [ ( f ( x ) + 2 ) 2 − 4 f ( x ) ] = 13 ( iii ) x → 2 lim [ f ( x ) − 2 ...

4.2

Jawaban terverifikasi